Given an array arr[] of size N and an integer K. The task is to find the last remaining element in the array after reducing the array. The rules for reducing the array are:
- The first and last element say X and Y are chosen and removed from the array arr[].
- The values X and Y are added. Z = X + Y.
- Insert the value of Z % K into the array arr[] at the position ((N/2) + 1)th position, where N denotes the current length of the array.
Examples:
Input: N = 5, arr[] = {1, 2, 3, 4, 5}, K = 7
Output: 1
Explanation:
The given array arr[] reduces as follows:
{1, 2, 3, 4, 5} -> {2, 6, 3, 4}
{2, 6, 3, 4} -> {6, 6, 3}
{6, 6, 3} -> {2, 6}
{2, 6} -> {1}
The last element of A is 1.Input: N = 5, arr[] = {2, 4, 7, 11, 3}, K = 12
Output: 3
Explanation:
The given array arr[] reduces as follows:
{2, 4, 7, 11, 3} -> {4, 5, 7, 11}
{4, 5, 7, 11} -> {5, 3, 7}
{5, 3, 7} -> {0, 3}
{0, 3} -> {3}
The last element of A is 3.
Naive approach: The naive approach for this problem is that at every step, find the first element and last element in the array and compute (X + Y) % K where X is the first element and Y is the last element of the array at every step. After computing this value, insert this value at the given position.
Time Complexity: O(N2)
Efficient Approach:
- On observing carefully, it can be said that the sum of the elements of the array modulo K is never changed in the array throughout.
- This is because we are basically inserting the value X + Y % K back into the array.
- Therefore, this problem can be solved in linear time by directly finding the sum of the array and finding the value sum % K.
Below is the implementation of the above approach:
C++
// C++ program to find the value of the // reduced Array by reducing the array // based on the given conditions #include <iostream> using namespace std; // Function to find the value of the // reduced Array by reducing the array // based on the given conditions int find_value(int a[], int n, int k) { // Variable to store the sum int sum = 0; // For loop to iterate through the // given array and find the sum for (int i = 0; i < n; i++) { sum += a[i]; } // Return the required value return sum % k; } // Driver code int main() { int n = 5, k = 3; int a[] = { 12, 4, 13, 0, 5 }; cout << find_value(a, n, k); return 0; } |
Java
// Java program to find the value of the // reduced Array by reducing the array // based on the given conditions public class GFG { // Function to find the value of the // reduced Array by reducing the array // based on the given conditions public static int find_value(int a[], int n, int k) { // Variable to store the sum int sum = 0; // For loop to iterate through the // given array and find the sum for (int i = 0; i < n; i++) { sum += a[i]; } // Return the required value return sum % k; } // Driver code public static void main(String[] args) { int n = 5, k = 3; int a[] = { 12, 4, 13, 0, 5 }; System.out.println(find_value(a, n, k)); } } |
Python3
# Python3 program to find the value of the # reduced Array by reducing the array # based on the given conditions # Function to find the value of the # reduced Array by reducing the array # based on the given conditions def find_value(a, n, k): # Variable to store the sum sum = 0 # For loop to iterate through the # given array and find the sum for i in range(n): sum += a[i] # Return the required value return sum % k # Driver code if __name__ == "__main__": n, k = 5, 3; a = [12, 4, 13, 0, 5]; print(find_value(a, n, k)) |
C#
// C# program to find the value of the // reduced Array by reducing the array // based on the given conditions using System; class GFG { // Function to find the value of the // reduced Array by reducing the array // based on the given conditions public static int find_value(int []a, int n, int k) { // Variable to store the sum int sum = 0; // For loop to iterate through the // given array and find the sum for (int i = 0; i < n; i++) { sum += a[i]; } // Return the required value return sum % k; } // Driver code public static void Main(string[] args) { int n = 5, k = 3; int []a = { 12, 4, 13, 0, 5 }; Console.WriteLine(find_value(a, n, k)); } } // This code is contributed by AnkitRai01 |
1
Time Complexity: O(N)
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Improved By : AnkitRai01
